Check whether Array can be made strictly increasing by shifting 1 value to the right
Last Updated :
14 Feb, 2022
Given an array arr[] of N positive integers, the task is to check whether an array can be made strictly increasing by shifting 1 value to its right element any number of times (i.e, for any value of i, decrement arr[i] and increment arr[i + 1] by 1).
Note: The integer at any index after any operation must not be negative.
Example:
Input: arr[] = {4, 5, 4}
Output: Yes
Explanation: The array can be made strictly increasing by performing the following operations:
- Choose i = 1 and shift value 1 from A1 to A2 . Now, arr[] = {3, 6, 4}.
- Choose i = 2 and shift value 1 from A2 to A3 . Now, arr[] = {3, 5, 5}.
- Choose i = 2 and shift value 1 from A2 to A3 . Now, arr[] = {3, 4, 6}.
Input: arr[] = {0, 1, 0}
Output: No
Approach: The given problem can be solved using a greedy approach. The idea is that for any index i, the least possible strictly increasing sequence without including the negative integers is 0, 1, 2, 3, … . Hence, for each value of i in the range [0, N), the sum of integers till arr[i] must be greater than the sum of the series 0, 1, 2, …, i-1 which is equivalent to (i * (i – 1))/2. Therefore, if the given array satisfies that condition, return “Yes“, otherwise, return “No“.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
bool isPossible(int arr[], int N)
{
int sum = 0;
for (int i = 0; i < N; i++) {
sum += arr[i];
if (sum < i * (i + 1) / 2)
return false;
}
return true;
}
int main()
{
int arr[] = { 4, 5, 4 };
int N = sizeof(arr) / sizeof(int);
if (isPossible(arr, N)) {
cout << "Yes";
}
else {
cout << "No";
}
return 0;
}
|
Java
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
static Boolean isPossible(int arr[], int N)
{
int sum = 0;
for (int i = 0; i < N; i++) {
sum += arr[i];
if (sum < i * (i + 1) / 2)
return false;
}
return true;
}
public static void main (String[] args) {
int arr[] = { 4, 5, 4 };
int N = arr.length;
if (isPossible(arr, N)) {
System.out.print("Yes");
}
else {
System.out.print("No");
}
}
}
|
Python3
def isPossible(arr, N):
sum = 0
for i in range(N):
sum += arr[i]
if sum < i * (i + 1) / 2:
return 0
return 1
arr = [4, 5, 4]
N = len(arr)
if isPossible(arr, N) == 1:
print("Yes")
else:
print("No")
|
C#
using System;
class GFG
{
static Boolean isPossible(int []arr, int N)
{
int sum = 0;
for (int i = 0; i < N; i++) {
sum += arr[i];
if (sum < i * (i + 1) / 2)
return false;
}
return true;
}
public static void Main () {
int []arr = { 4, 5, 4 };
int N = arr.Length;
if (isPossible(arr, N)) {
Console.Write("Yes");
}
else {
Console.Write("No");
}
}
}
|
Javascript
<script>
function isPossible(arr, N) {
let sum = 0;
for (let i = 0; i < N; i++) {
sum += arr[i];
if (sum < i * (i + 1) / 2)
return false;
}
return true;
}
let arr = [4, 5, 4];
let N = arr.length;
if (isPossible(arr, N)) {
document.write("Yes");
}
else {
document.write("No");
}
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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